Question

When a ceiling fan is switched off its angular velocity reduces to half of its value, while it makes $12$ rotations. How many more rotations will it make before coming to rest? (Assume uniform angular retardation)

• A. $18$
• B. $24$
• C. $4$
• D. $24$

Answer 1

The correct option is C $4$

In the situation $\left(\text{I}\right)-$ It is given that angular velocity reduces to half of it's value and making $12$ rotations

By the equation of motion under constant angular acceleration,
${\omega }^2={\omega }_0^2+2\alpha {\theta }_1$
also given that angular velocity reduces to half of it's value so, $\omega =\frac{{\omega }_0}{2}$ and $\theta =2\pi n$ (where, $n=12$)
$\theta =2\times 12\pi =24\pi$

$\Rightarrow \frac{{\omega }_0^2}{4}={\omega }_0^2+2\alpha \times 24\pi$
$48\pi \times \alpha =-\frac{3{\omega }_0^2}{4}\Rightarrow \alpha =-\frac{{\omega }_0^2}{64\pi }$

Now in the situation $\text{II}-$
It is said that the fan comes to rest, hence $(\omega =0)$ and the initial angular velocity here will be $\left(\frac{{\omega }_0}{2}\right)$ as the angular acceleration is constant we get $(\alpha =-\frac{{\omega }_0^2}{64\pi })$, applying this in the equation we get,

$0=\frac{{\omega }_0^2}{4}+2\times -\frac{{\omega }_0^2}{64\pi }\times {\theta }_2$
$\Rightarrow {\theta }_2=8\pi$,
here ${\theta }_2$ is the angular displacement before it comes to rest again

Hence the number of rotation before it comes to rest is $\frac{{\theta }_2}{2\pi }=\frac{8\pi }{2\pi }=4$

Hence option C is the correct answer.

(OR)

By the equation of motion under constant angular acceleration,
${\omega }^2={\omega }_0^2+2\alpha \theta$
Here $\theta =2\pi n$ where $n$ is number of rotations.

Case -$\left(I\right)$
It is given that angular velocity reduces to half of it's value and making $12$ rotations.

$(\frac{{\omega }_0}{2}{)}^2={\omega }_0^2+2\alpha (2\pi \left(12\right))$
$(\frac{{\omega }_0}{2}{)}^2-{\omega }_0^2=2\alpha (2\pi \left(12\right))$
$-\frac{3{\omega }_0^2}{4}=2\alpha \left(2\pi \right(12\left)\right).............\left(1\right)$

Case-$\left(II\right)$

It is said that the fan comes to rest, hence $(\omega =0)$ and the initial angular velocity here will be $\left(\frac{{\omega }_0}{2}\right)$.
Let $n$ be number of rotation before coming to rest.

$0^2=(\frac{{\omega }_0}{2}{)}^2+2\alpha (2\pi \left(n\right))$
$-(\frac{{\omega }_0}{2}{)}^2=2\alpha (2\pi \left(n\right))......(2)$

Dividing equations $\left(1\right)$ and $\left(2\right)$
$\frac{3}{1}=\frac{12}{n}$
$n=4$